Question

The value of observed and calculate molecular weight of calcium nitrate are respectively $65.6$ and $164.$ the degree of dissociation of calcium nitrate is
a) $25\% $
b) $50\% $
c) $75\% $
d) $60\% $

Answer 1

Hint:The relation of Van't Hoff factor$(i)$ and degree of dissociation $(\alpha )$will be used. Knowledge about colligative properties and use of Van't Hoff factor$(i)$ should be kept in mind. There are different formulas used to calculate Van't Hoff factor$(i)$.

Complete step by step solution:
Van't Hoff factor$(i)$ is the used to calculate effect of solute on colligative properties. It is the ratio of actual molecular weight and observed calculated weight of a compound.
For strong electrolytes, Van't Hoff factor$(i)$ will be equal to the number of total ions it is giving into solution after complete dissociation. For non-electrolytes, the value of Van't Hoff factor$(i)$ taken as one.
When we add a solute in a particular solution either it can react with a solution or it can not react, but if it reacts, it can either undergo association or dissociation.
This is a particular case of dissociation of a compound in solution.
Now the dissociation will be:
$Ca{(N{O_3})_2} \to C{a^{2 + }} + 2NO_3^ - $
It is given that calculated molecular weight is $164.$ and observed molecular weight is$65.6$.
As \[Vant{\text{ }}Hoff{\text{ }}factor\;(i) = \dfrac{{calculated - molecular wt.}}{{observed - molecular wt.}}\] ,
$
 \Rightarrow i = \dfrac{{164}}{{65.6}} \\
 \Rightarrow i = 2.5 \\
$
For the reaction: $Ca{(N{O_3})_2} \to C{a^{2 + }} + 2NO_3^ - $

Initial concentration:	$1$	$ 0$	$ 0$
Final concentration:	$1 - \alpha $	$\alpha $	$2\alpha $

Total concentration will be $i = 1 - \alpha + \alpha + 2\alpha $
$ \Rightarrow i = 1 + 2\alpha $
Solving for $\alpha $ we get,
$ \Rightarrow \alpha = \dfrac{{i - 1}}{2}$
By substituting the value of $i$ in the above equation we get,
$
\Rightarrow \alpha = \dfrac{{2.5 - 1}}{2} \\
\Rightarrow \alpha = \dfrac{{1.5}}{2} \\
\therefore \alpha = 0.75 \\
$
As the degree of dissociation $\alpha = 0.75$ , the $\% $of degree of dissociation is $75\% $.
$\therefore $ Degree of dissociation for Calcium nitrate will be $75\% $.
Hence,the correct option is (C).

Additional Information: There are different formulas that we can use to calculate the Van't Hoff factor$(i)$, but here this formula is directly applicable in this case.


Note:Solute in a solution can undergo either association or dissociation. For e.g. in case of acetic acid in benzene solvent. Acetic acid in benzene solvent actually exists as a dimer Hence the value of Van't Hoff factor$(i)$ will be equal to$2$.